I know that $\mathbb{R}^4$ admits uncountably many differentiable structures and I was wandering what happen if we consider closed (or just compact) 4-manifolds. Are there any closed (or compact) 4-manifolds with uncountably many differentiable structures? And with countably many?

  • $\begingroup$ Yes, I want to know if there are any uncountable (or countable) family of closed (or compact) 4-manifolds which are homeomorphic but not diffeomorphic. $\endgroup$
    – Dario
    Jun 14, 2014 at 14:21
  • 1
    $\begingroup$ Donaldson showed that Dolgachev surfaces (en.wikipedia.org/wiki/Dolgachev_surface) have countably many smooth structures. $\endgroup$ Jun 14, 2014 at 17:38
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    $\begingroup$ Answered at mathoverflow.net/questions/171978/… in a comment. $\endgroup$ Jun 17, 2014 at 17:41

1 Answer 1


This is a summary of comments.

Your question was answered here in a comment by Ryan Budney. In the nutshell: The categories DIFF and PL are equivalent in dimension 4 (this is due to Kirby and Siebenmann, but one can easily spend a year trying to understand their proof). A PL structure on an n-manifold is equivalent to a "combinatorial triangulation", i.e., a triangulation where the link of each simplex is combinatorially homeomorphic to the closed n-ball. Since there are only countably many finite simplicial complexes, it follows that there are only countably many closed differentiable 4-manifolds.

Furthermore, I think, it is also true in higher dimensions that there are only countably many smooth closed manifolds (even though, DIFF $\ne$ PL), but I would have to check this.

Lastly, according to Qiaochu Yuan's comment, Dolgachev surface (which is a closed 4-manifold) admits countable infinitely many smooth structures.


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